## q=-q question (7.39)

martha-1I
Posts: 76
Joined: Fri Sep 26, 2014 2:02 pm

### q=-q question (7.39)

When asked to solve for the final temperature of "A 50g ice cube at 0$\circ$C [when] added to a glass containing 400g of water at 45$\circ$C" (7.39), I plugged in mC$\bigtriangleup$T=-mC$\bigtriangleup$T to find Tf. I ended up with the answer 42.4$\circ$C, but the back of the book has the answer as 31$\circ$C. I noticed from the solution manual that the grams where turned into moles and multiplied by 6.01*103J/mol-1 and added to mC$\bigtriangleup$T to calculate the heat of the ice cube. Why is it that the grams were turned into moles and were multiplied by 6.01*103 then they were added to mC$\bigtriangleup$T?

Chem_Mod
Posts: 18881
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

### Re: q=-q question (7.39)

Notice that in this scenario, some ice cubes are being put into a relatively larger quantity of warm water. A phase change will occur. Remember that a phase change is associated with an enthalpy, in this case, the enthalpy of fusion. The heat transferred is first being used to melt the ice. Only after it's all melted will the heat transfer result in an equilibration of temperature.

Therefore, all you need is to add $m_{ice} \Delta H_{fus}$ to the left side of your equation. Of course, the units must be consistent, so if the enthalpy is given in the problem as J/mol, then you have to convert the grams to moles. 31 degrees Celsius is correct for this problem.

Alternatively, if you don't want to convert anything to moles, and want to stick with grams, you can check that using 334 J/g for the enthalpy of fusion and 4.184 J/g*C for the specific heat of water gives the same answer too.